# Ultra basic classical physics: force, energy, power

March 22, 2010 1:14 PM Subscribe

Please help me understand the relationship between force and power via this little thought experiment I dreamed up.

I'm just doing a little bit of basic review of classical physics, because I've been thinking about a hard science fiction video game, and I can't wrap my head around the difference between force and energy.

I think the best way to describe my confusion is with a little thought-experiment I've been playing.

Okay, so you and I live on different planets, and they're very close, and the path from our planets to our respective moons are parallel, and in the same direction, which we'll call east. But my planet, and my moon, is moving west at a rate of 1 meter per second in relation to your planet and your moon.

We have the same model of spaceship. It weighs 1 kg, and is powered by a super high-tech nuclear reactor with a maximum power output of 1.5 watt.

I begin a trip from my planet to my moon. I quickly accelerate to 2 meters per second, because I'm in a hurry. You also begin a trip to your moon, but you're not in a hurry, so you only go at 1 meter per second. And we discover that we're traveling right next to each other. Relative to my source and destination, we're both going 2m/s; relative to yours, we're both going 1m/s; and relative to each other, we're stationary.

We decide to race a little bit, and each step on the pedal. For you to reach 2m/s from your 1m/s in 1 second, that means 1 newton applied over a distance of 1.5 meters, right? So that's 1.5 joules, over 1 second, or 1.5 watt, and your reactor can handle it. But for me, that's a change from 2m/s to 3m/s over a distance of 2.5 meters in 1 second. That's 2.5 watts, and my little reactor can't handle it. Relative to your trip, I fall behind.

What am I missing? Is this really the way it works? At the least, it is extraordinarily counter-intuitive.

I'm just doing a little bit of basic review of classical physics, because I've been thinking about a hard science fiction video game, and I can't wrap my head around the difference between force and energy.

I think the best way to describe my confusion is with a little thought-experiment I've been playing.

Okay, so you and I live on different planets, and they're very close, and the path from our planets to our respective moons are parallel, and in the same direction, which we'll call east. But my planet, and my moon, is moving west at a rate of 1 meter per second in relation to your planet and your moon.

We have the same model of spaceship. It weighs 1 kg, and is powered by a super high-tech nuclear reactor with a maximum power output of 1.5 watt.

I begin a trip from my planet to my moon. I quickly accelerate to 2 meters per second, because I'm in a hurry. You also begin a trip to your moon, but you're not in a hurry, so you only go at 1 meter per second. And we discover that we're traveling right next to each other. Relative to my source and destination, we're both going 2m/s; relative to yours, we're both going 1m/s; and relative to each other, we're stationary.

We decide to race a little bit, and each step on the pedal. For you to reach 2m/s from your 1m/s in 1 second, that means 1 newton applied over a distance of 1.5 meters, right? So that's 1.5 joules, over 1 second, or 1.5 watt, and your reactor can handle it. But for me, that's a change from 2m/s to 3m/s over a distance of 2.5 meters in 1 second. That's 2.5 watts, and my little reactor can't handle it. Relative to your trip, I fall behind.

What am I missing? Is this really the way it works? At the least, it is extraordinarily counter-intuitive.

Sorry, this isn't clear. Please replace my second paragraph with the following:

To enter the same frame of reference, you had to accelerate twice as long as your friend, but you are now at

posted by bonehead at 1:41 PM on March 22, 2010

To enter the same frame of reference, you had to accelerate twice as long as your friend, but you are now at

**equal speeds in both external frames of reference**. If both you and they accelerate for 1s then you will still both have the same**relative**speed**(that is, zero)**, but 1.5 m/s faster to**either of**your external frames of reference.posted by bonehead at 1:41 PM on March 22, 2010

Because energy, and hence power, is proportional not just to force, but to distance. My problem is that the distance being measured depends on the frame of reference.

From the frame of reference of my planet and my moon, I am changing my velocity from 2m/s to 3m/s. Over that time, I have an average velocity of 2.5 m/s. In that second that I accelerate, I cover 2.5 meters. At a force of 1 newton, since I weigh 1 kg, I require 2.5 joules, and 2.5 watts.

From the frame of reference of your planet and your moon, I am changing my velocity from 1m/s to 2m/s. Over that time, I have an average velocity of 1.5m/s. I cover 1.5 meters. At a force of 1 newton, I require 1.5 joules and 1.5 watts.

And, if you want, from the frame of reference of you and me, I am changing my velocity from 0m/s to 1m/s. I cover 0.5m. I require 0.5 watts.

I'm sure there's something very basic that I'm not understanding.

posted by nathan v at 2:03 PM on March 22, 2010

From the frame of reference of my planet and my moon, I am changing my velocity from 2m/s to 3m/s. Over that time, I have an average velocity of 2.5 m/s. In that second that I accelerate, I cover 2.5 meters. At a force of 1 newton, since I weigh 1 kg, I require 2.5 joules, and 2.5 watts.

From the frame of reference of your planet and your moon, I am changing my velocity from 1m/s to 2m/s. Over that time, I have an average velocity of 1.5m/s. I cover 1.5 meters. At a force of 1 newton, I require 1.5 joules and 1.5 watts.

And, if you want, from the frame of reference of you and me, I am changing my velocity from 0m/s to 1m/s. I cover 0.5m. I require 0.5 watts.

I'm sure there's something very basic that I'm not understanding.

posted by nathan v at 2:03 PM on March 22, 2010

The work done in each case is the difference in kinetic energy from the initial state to the final state. Kinetic energy goes up quadratically with speed. As far as work is concerned, it doesn't matter whether it's a large force over a small distance or a small force over a long distance. For a classical case such as this it's this simple:

For person 1 (i=initial, f=final)

KEi = 1/2 * m1,i * v1,i^2 = 0.5*1*1^2=0.5J

KEf = 1/2 * m1,f * v1,f^2 = 0.5*1*2^2=2J

W=KE2-KE1 = 1.5J

P=W/delta_t=1.5W

For person 2 (i=initial, f=final)

KEi = 1/2 * m2,i * v2,i^2 = 0.5*1*2^2=2J

KEf = 1/2 * m2,f * v2,f^2 = 0.5*1*3^2=4.5J

W=KE2-KE1 = 2.5J

P=W/delta_t=2.5W

There are some exceptions to the large force/small force thing. Very large forces can induce localized deformation (e.g. bullet hitting a flak jacket) and funny things happen when time and/or length scales are changed but for your "ultra basic" example, the difference is just that kinetic energy goes up faster than velocity itself.

posted by KevCed at 2:29 PM on March 22, 2010

For person 1 (i=initial, f=final)

KEi = 1/2 * m1,i * v1,i^2 = 0.5*1*1^2=0.5J

KEf = 1/2 * m1,f * v1,f^2 = 0.5*1*2^2=2J

W=KE2-KE1 = 1.5J

P=W/delta_t=1.5W

For person 2 (i=initial, f=final)

KEi = 1/2 * m2,i * v2,i^2 = 0.5*1*2^2=2J

KEf = 1/2 * m2,f * v2,f^2 = 0.5*1*3^2=4.5J

W=KE2-KE1 = 2.5J

P=W/delta_t=2.5W

There are some exceptions to the large force/small force thing. Very large forces can induce localized deformation (e.g. bullet hitting a flak jacket) and funny things happen when time and/or length scales are changed but for your "ultra basic" example, the difference is just that kinetic energy goes up faster than velocity itself.

posted by KevCed at 2:29 PM on March 22, 2010

Right, that's how I read it too, but it leads to the situation I described: 2 identical spaceships, at no velocity relative to each other, but one falls behind. Doesn't it? Since your person 2 requires 2.5 watts, and your person 1 requires 1.5 watts.

posted by nathan v at 2:47 PM on March 22, 2010

posted by nathan v at 2:47 PM on March 22, 2010

Having not taken physics in a few years, I found this an interesting question. I

Therefore, in order for any comparison between your spaceships' relative power outputs to be valid, they need to be measured from the same frame of reference. One spaceship will not fall behind because a given amount of power will result in both of them accelerating from the same initial velocity to the same target velocity as long as you hold the frame of reference constant. Holding the frame of reference constant is essential if you're going to be comparing quantities (velocity, kinetic energy, power) between the ships because otherwise you're not comparing the same thing.

posted by stufflebean at 3:01 PM on March 22, 2010

*think*the key to understanding this is that all of the quantities you're interested have no real meaning without a frame of reference. It's easy to see why this is counter-intuitive. My car is rated at 165 horsepower, and we don't talk about the frame of reference for which this holds, but this power rating only really holds for a very particular set of assumptions (the frame of reference is the ground I'm travelling over, etc.). If we were to measure the power of my car from another frame of reference, the power would be different, even though nothing has occurred to change my car. Therefore, my car isn't really a "165 HP car" (and your ships don't really have a "1.5 W engine") but rather it's a "165 HP car given this set of conditions: { ... }."Therefore, in order for any comparison between your spaceships' relative power outputs to be valid, they need to be measured from the same frame of reference. One spaceship will not fall behind because a given amount of power will result in both of them accelerating from the same initial velocity to the same target velocity as long as you hold the frame of reference constant. Holding the frame of reference constant is essential if you're going to be comparing quantities (velocity, kinetic energy, power) between the ships because otherwise you're not comparing the same thing.

posted by stufflebean at 3:01 PM on March 22, 2010

Sorry, should be "... the key to understanding this is that all of the quantities you're interested

Should have previewed from a different frame of reference.

posted by stufflebean at 3:09 PM on March 22, 2010

__have no real meaning ..."__*in*Should have previewed from a different frame of reference.

posted by stufflebean at 3:09 PM on March 22, 2010

Don't forget that both spaceships also have to satisfy conservation of momentum. This means that you have to push something (rocket fuel) in the opposite direction of your acceleration. So, in your example, let's say both spaceships also have 0.1 kg of rocket fuel. To accelerate from 1 m/s to 2 m/s, conservation of momentum requires:

m

(1 kg)(1 m/s) + (0.1 kg)(1 m/s) = (1 kg)(2 m/s) + (0.1 kg)(-9 m/s)

If you calculate the kinetic energy: 1/2 * (m

initial: 1/2 * ( (1 kg)*(1 m/s)^2 + (0.1 kg)*(1 m/s)^2) = 0.55 J

final: 1/2*( (1 kg)*(2 m/s)^2 + (0.1 kg)*(-9 m/s)^2) = 6.05 J

difference:

Now, if you repeat the calculation with an initial velocity of 2 m/s, accelerating to 3 m/s, conservation of momentum requires:

(1 kg)(2 m/s) + (0.1 kg)(2 m/s) = (1 kg)(3 m/s) + (0.1 kg)(-8 m/s)

and the kinetic energy is:

initial: 1/2 * ( (1 kg)*(2 m/s)^2 + (0.1 kg)*(2 m/s)^2) = 2.2 J

final: 1/2 *( (1 kg)*(3 m/s)^2 + (0.1 kg)*(-8 m/s)^2) = 7.7 J

difference:

So, the change in kinetic energy needed for each ship to increase its velocity by 1 m/s is independent of the frame of reference.

posted by yarmond at 3:18 PM on March 22, 2010

m

_{1,i}*v_{1,i}+ m_{2,i}*v_{2,i}= m_{1,f}*v_{1,f}+ m_{2,f}*v_{2,f}(1 kg)(1 m/s) + (0.1 kg)(1 m/s) = (1 kg)(2 m/s) + (0.1 kg)(-9 m/s)

If you calculate the kinetic energy: 1/2 * (m

_{1}*v_{1}^{2}+ m_{2}*v_{2}^{2})initial: 1/2 * ( (1 kg)*(1 m/s)^2 + (0.1 kg)*(1 m/s)^2) = 0.55 J

final: 1/2*( (1 kg)*(2 m/s)^2 + (0.1 kg)*(-9 m/s)^2) = 6.05 J

difference:

**5.5 J**Now, if you repeat the calculation with an initial velocity of 2 m/s, accelerating to 3 m/s, conservation of momentum requires:

(1 kg)(2 m/s) + (0.1 kg)(2 m/s) = (1 kg)(3 m/s) + (0.1 kg)(-8 m/s)

and the kinetic energy is:

initial: 1/2 * ( (1 kg)*(2 m/s)^2 + (0.1 kg)*(2 m/s)^2) = 2.2 J

final: 1/2 *( (1 kg)*(3 m/s)^2 + (0.1 kg)*(-8 m/s)^2) = 7.7 J

difference:

**5.5 J**So, the change in kinetic energy needed for each ship to increase its velocity by 1 m/s is independent of the frame of reference.

posted by yarmond at 3:18 PM on March 22, 2010

Wow, people are making this way more difficult than it needs to be.

As an example, if I'm pushing a crate up a ramp

Your numbers only coincidentally work out in the case of the other spaceship due to so many of the numbers being 1.

posted by 0xFCAF at 3:30 PM on March 22, 2010 [1 favorite]

**The**The underlined calculation is where you went astray:`energy = force * distance`equation only applies when the distance being covered is due to the force being applied!*For you to reach 2m/s from your 1m/s in 1 second, that means 1 newton applied over a distance of 1.5 meters, right? So that's 1.5 joules, over 1 second, or 1.5 watt, and your reactor can handle it. But for me, that's a change from 2m/s to 3m/s over a distance of 2.5 meters in 1 second.*__That's 2.5 watts__, and my little reactor can't handle it. Relative to your trip, I fall behind.As an example, if I'm pushing a crate up a ramp

*and*I happen to be on the Earth which is traveling at hundreds of miles per hour (or whatever) through space in the same direction as I'm pushing the crate, by the time I'm done the crate will have moved thousands of miles, but I definitely didn't do billions of joules of work. The distance covered by the Earth's movement is not a factor in the work calculation; similarly the distance covered by the spaceship's*initial*acceleration is not a factor in the work calculation of the*later*acceleration.Your numbers only coincidentally work out in the case of the other spaceship due to so many of the numbers being 1.

posted by 0xFCAF at 3:30 PM on March 22, 2010 [1 favorite]

Work is a non-useful quantity in this situation. The Impulse is much easier to think about. The rocket engine can provide a certain amount of impulse by shooting material out the other direction. The net momentum change of the rocket+fuel system is zero (equal and opposite, you know), and there's no need for any external reference frame to confuse us.

When you invoke the idea of work,

posted by kiltedtaco at 4:02 PM on March 22, 2010

When you invoke the idea of work,

**you must be doing work on something**. In this case it's the rocket fuel (or the fuel is doing work on the rocket), which is why you're having so much trouble since it too is moving. Both rockets do the same amount of work on the fuel that they eject, and thus they can both accelerate at the same rate.posted by kiltedtaco at 4:02 PM on March 22, 2010

You are switching frames of reference mid-calculation, and that is messing you up.

Ok.

No, you've just switched reference frames. In the frame where I am going 1m/s (say, the frame of an observer on my planet) you are

posted by abc123xyzinfinity at 4:26 PM on March 22, 2010

*For you to reach 2m/s from your 1m/s in 1 second, that means 1 newton applied over a distance of 1.5 meters, right? So that's 1.5 joules, over 1 second, or 1.5 watt, and your reactor can handle it.*Ok.

*But for me, that's a change from 2m/s to 3m/s over a distance of 2.5 meters in 1 second.*No, you've just switched reference frames. In the frame where I am going 1m/s (say, the frame of an observer on my planet) you are

*also*going 1m/s (the relative speed between you and the observer on my planet), so you would do the same calculation for your work energy. In fact, the way you've set up the problem, you and I are at rest with respect to one another in*all*frames of reference, so we would naturally agree upon the amount of power required to keep one another apace.posted by abc123xyzinfinity at 4:26 PM on March 22, 2010

*you and I are at rest with respect to one another in all frames of reference*

Well, all non-rotating frames anyway.

posted by abc123xyzinfinity at 4:29 PM on March 22, 2010

The trouble with explaining stuff like this is that there are many different-looking but equivalent paths to the right answer. Virtually everyone above has given a valid explanation, but it looks like they're arguing because one person is talking about how you switched reference frames, the next is explaining that W = F x d only works if the F is the only cause of the d, etc. -- 0xFCAF, kiltedtaco, abc123xyzinfinity, and yarmond, for example, are all making the same physical argument.

If you're trying to get a better working of understanding of physics, wrestling with

posted by range at 6:42 PM on March 22, 2010

If you're trying to get a better working of understanding of physics, wrestling with

*that*might be a good place to start. The simplest "dual-path" example I know of is projectile motion, which you can understand equally well via force (for a gun pointed straight up, what is the initial velocity, what is the decelerating force due to gravity, therefore what is the max height) or energy (initial velocity & mass gives you starting kinetic energy, which will be fully converted to potential energy at the max height).posted by range at 6:42 PM on March 22, 2010

I think the reason you're confused is because it doesn't make any sense to characterize a rocket motor based on its power output, like kiltedtaco says. A rocket motor works by pushing out mass at a given flow rate and velocity relative to the nozzle. If both of those are constant (and neglecting the decrease of mass in the ship due to fuel) then what results is a constant force of X newtons. Now consider two different frames of reference of that same identical rocket with its motor running at the same output. Say in one frame the rocket appears to be moving at 5 m/s and so the power output of the motor is 5*X watts. In another frame the rocket might appear to only be moving 3 m/s and so the power output of the motor is 3*X watts -- and remember from the standpoint of the people on the ship, this motor is operating identically. So what would we say the motor is rated for, 3*X watts or 5*X watts? We can't, because the rate at which the motor appears to be increasing the kinetic energy of the rocket depends on how we choose the frame of reference for observing that rocket. Our rulers need to have zero at the same mark.

posted by Rhomboid at 1:15 AM on March 23, 2010

posted by Rhomboid at 1:15 AM on March 23, 2010

Thank you all so much for your help. I understand now. And it works exactly how I feel that it SHOULD work. (Except I wasn't really anticipating this relationship between energy and reaction mass/velocity, which is really interesting to me, and I'm exploring it right now.)

I worked very hard on a reformulation of the experiment, without switching frames (except to check the work) and I want to share it, in case its useful to anybody, even though I think I understand how it works now.

So:

I live in a 1-dimensional universe, where everything's an intangible point entity. This universe has two planets, which at time 0 occupy the same space, and a moon, which at time 0 is 1.5 meters from the planets. Planet A has a velocity of -1m/s relative to both planet B and planet A. I leave planet A in my spaceship, with a reactor capable of 1.5W peak power, and super thrusters capable of 1m/s/s acceleration, headed to Moon. Can I get there in three seconds if I floor it?

From the frame of planet A: In the first second, I cover 0.5m and accelerate to 1m/s, under 0.5W of power, while Moon grows 1m more distant. In the next second, I cover 1.5m and accelerate to 2m/s, under 1.5W of power, while Moon grows 1m more distant. The moon is now 1.5 meters away, and falling away, but I'm getting faster. In the third second, I try my hardest to reach 3m/s, which would move me 2.5m this second, enough to reach the moon and compensate for its velocity-- but my reactor sputters. Reaching that speed would require 2.5W, totally out of reach.

From the frame of planet B: In the first second, I accelerate from -1m/s to 0 m/s, but actually lose 0.5 m, while the moon stays still. In the second second, I gain back that 0.5m, accelerating to 1m/s. The moon is still 1.5 meters away! I step on it. I accelerate to 2m/s, which covers 1.5m. Can my reactor handle it? Woot, it can.

posted by nathan v at 1:18 AM on March 23, 2010

I worked very hard on a reformulation of the experiment, without switching frames (except to check the work) and I want to share it, in case its useful to anybody, even though I think I understand how it works now.

So:

I live in a 1-dimensional universe, where everything's an intangible point entity. This universe has two planets, which at time 0 occupy the same space, and a moon, which at time 0 is 1.5 meters from the planets. Planet A has a velocity of -1m/s relative to both planet B and planet A. I leave planet A in my spaceship, with a reactor capable of 1.5W peak power, and super thrusters capable of 1m/s/s acceleration, headed to Moon. Can I get there in three seconds if I floor it?

From the frame of planet A: In the first second, I cover 0.5m and accelerate to 1m/s, under 0.5W of power, while Moon grows 1m more distant. In the next second, I cover 1.5m and accelerate to 2m/s, under 1.5W of power, while Moon grows 1m more distant. The moon is now 1.5 meters away, and falling away, but I'm getting faster. In the third second, I try my hardest to reach 3m/s, which would move me 2.5m this second, enough to reach the moon and compensate for its velocity-- but my reactor sputters. Reaching that speed would require 2.5W, totally out of reach.

From the frame of planet B: In the first second, I accelerate from -1m/s to 0 m/s, but actually lose 0.5 m, while the moon stays still. In the second second, I gain back that 0.5m, accelerating to 1m/s. The moon is still 1.5 meters away! I step on it. I accelerate to 2m/s, which covers 1.5m. Can my reactor handle it? Woot, it can.

posted by nathan v at 1:18 AM on March 23, 2010

"Planet A has a velocity of -1m/s relative to both planet B and planet A"

erm, that would be: "..relative to both Planet B and Moon."

posted by nathan v at 1:27 AM on March 23, 2010

erm, that would be: "..relative to both Planet B and Moon."

posted by nathan v at 1:27 AM on March 23, 2010

To echo what's above, work and energy are frame dependent. Why wouldn't they be? Momentum is. To answer your question you need to figure out what you mean when you say that an engine "is powered by a super high-tech nuclear reactor with a maximum power output of 1.5 watt." This is defined in a particular frame. If you mean in the rest frame of the spaceship, then it relates directly to the thrust of the engines. Thrust (being a force) will be independent of inertial frame.

Please do not listen to the comment by 0xFCAF. It will only lead to confusion in other problems. For example, one might conclude that you could fire a thruster on a satellite for a fixed amount of time either parallel or perpendicular to the line of motion and end up with the same final orbital energy. This is incorrect. In his example, the work done is indeed "billions of joules" in

posted by dsword at 10:56 AM on March 23, 2010

Please do not listen to the comment by 0xFCAF. It will only lead to confusion in other problems. For example, one might conclude that you could fire a thruster on a satellite for a fixed amount of time either parallel or perpendicular to the line of motion and end up with the same final orbital energy. This is incorrect. In his example, the work done is indeed "billions of joules" in

*some*inertial frame. But this says nothing about the amount of effort he had to expend.posted by dsword at 10:56 AM on March 23, 2010

This thread is closed to new comments.

To enter the same frame of reference, you had to accelerate twice as long as your friend, but you are now in the same frame of reference. If both you and they accelerate for 1s then you will still both have the same speed, but 1.5 m/s faster to your external frames of reference.

You will be approaching your moon at 2 + 1.5 = 3.5 m/s, they will be approaching theirs at 1 + 1.5 = 2.5 m/s.

Unless I'm misunderstanding, this:

But for me, that's a change from 2m/s to 3m/s over a distance of 2.5 meters in 1 second, is where you're having you're problem. This isn't true in your mutual frame of reference (how fast you are going relative to each other).posted by bonehead at 1:37 PM on March 22, 2010